Physics question thread

[Shrike]

All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

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[5moreminutes]

A particle moving at 10 m/s reverses its direction to move at 20 m/s in the opposite direction. If its acceleration is -10 m/s^2. what is the total distance that it travels...

I just dont understand the question being asked..can someone please clarify..this is # 61 from EK~
Thanks!

 

[dill_punjabi]

If it's acceleration is -10m/s2, and initial speed 10 m/s, then distance traveled in that particular direction will be 10 m.
And now it's changes it direction and travel with 20 m/s, and again acceleration -10m/s2, now distance during 1st second 20 m and it velocity reduced to 10 m/s. In the next second it will travel another 10 m before coming to rest.
So Total Distance 10 + 20 + 10 = 40 m
Total Displacement = 20


QQQ:- Two Vectors have unequal magnitudes. Can their sum be zero? yes/no(explain)
QQQ:- Two projectiles are thrown with same initial speed. One at angel x with respect to the level ground and other one at angel 90 -x. Both projectiles strike the ground at the same distance from projection point. Are both projectiles in the air for the same length of time?

 

[5moreminutes]

If it's acceleration is -10m/s2, and initial speed 10 m/s, then distance traveled in that particular direction will be 10 m.
And now it's changes it direction and travel with 20 m/s, and again acceleration -10m/s2, now distance during 1st second 20 m and it velocity reduced to 10 m/s. In the next second it will travel another 10 m before coming to rest.
So Total Distance 10 + 20 + 10 = 40 m
Total Displacement = 20


QQQ:- Two Vectors have unequal magnitudes. Can their sum be zero? yes/no(explain)
QQQ:- Two projectiles are thrown with same initial speed. One at angel x with respect to the level ground and other one at angel 90 -x. Both projectiles strike the ground at the same distance from projection point. Are both projectiles in the air for the same length of time?

Kiddhan dill_punjabi,..
thanks for the reply..well I was thinking the same way..but somehow the answer is 25 m is the total distance covered. they are not asking for displacement,but the displacement...
thanks

 

[Playmakur42]

If you take any object at rest and it explodes, why is total momentum zero immediately after the explosion? I know P=mv

My line of thinking is this: Each fragment has a mass and a (very high) velocity after the explosion, so how can it not have a momentum?????

Anyone know why?

 

[gridiron]

I am having a difficult time with Newton's Law of Universal Gravitation. Is there any one that can help me figure out this problem: I understand the problem but need to understand how to use the ratios that are involved. Can someone show me as it applies to these two problems.

1. Planet B has twice the mass of Planet A. Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10 m above the surface of Planet B and at the same time a 10 kg mass is dropped 10m above the surface of Planet A. If the mass on Planet B strikes the ground in 10 sec,the mass on Planet A strikes the ground in approx:
a.7, b. 10, c. 14, d.20 ans is a.
2. Planet A and B have the same mass . Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10m above the surface of Planet A and at the same a 5kg mass is dropped 10m above the surface of Planet B. If the mass on Planet B strikes the ground in 10 sec, the mass of Planet A strikes the ground in :
a. 2.5s b.5s c.10s d. 20s ans is b.

thanks

Hey! The first thing you need to know is the equation for Newton's law of gravitation. Newton's law of gravitation states that two objects exert a gravitational force of attraction on each other. The equation is:

F = Gm1m2/r^2

Where G is Newton's constant, don't worry about knowing that for the MCAT, m1 is the mass of object 1, m2 is mass of object 2 and r is the distance along the line of the two objects--m1 and m2.

So, how do you find the force of gravity near the planets surface? Given some object of m, you can manipulate Newton's law of gravitation. Suppose you want to find the gravitational force of attraction between the object and the planet. Say the mass of the object is m1 and the mass of the planet is m2. The force of gravity through the center of mass of any object on any planet is given by: F=mg. So the force of gravity on the object will be m1g. You can set the force of gravity equation equal to the law of gravitation because the describe the same thing. What you will be left with is:

g = GM(planet)/r^2(planet).

This way, given the appropriate numbers, you can solve the value of g for any planet. Another thing you need to recognize is that you can use this equations with ratios. The above equation for earth will always equal to 9.8 m/s^2 unless otherwise told. So if you have the dimensions of another planet with respect to earth, you can approximate the gravitational force on that planet with respect to earth using algebraic manipulation.

Another point worth noting before moving on: the mass of the object has no effect over the time it falls when the gravitational force is considered. What does this mean? Take earth for example. If you have two balls, of different mass, and throw them off a 100 meter building, both will fall with the same time. If you were considering another planet, mass of the ball and time it falls is independent. Now to solve your first question...

1.) The best way to attack any physics problem is to list the knows and unknowns. I don't suggest this for the MCAT, but when you are first solving physics problems it will help to realize the information you are given and what you are supposed to find. Through practice, you will be able to simplify the steps involved.

Knowns: distance the object falls for planet A and B, time it takes for object to fall for planet B, Mass of B with respect to A, radius of A with respect to B.

Unknowns: force of gravity of A and B, time it takes for object to fall for planet A.

What you first need to recognize using the knowns provided, is that you are going to need to use kinematic equations to solve for the accleration due to gravity on planet B. The equation you will use, provided zero initial velocity, is:

y = 0.5aT^2.

For planet B, y=10m, t=10 sec, so you obtain the acceleration due to gravity to be 0.2m/s^2. Now what? You need to take the ratio of gB to gA. You do this because you know gB. When you do this using the above equation you get:

gB/gA = MB/MA * (rA^2/rB^2).

What else are you told? From the question you are told:

MA = 0.5MB
rA = 0.5rB

If you plug in the following relations you are left with:

gB/gA = 0.5 -----> gB = 0.5gA.

This is the relation you seek. Thus, gA = 0.4 m/s^2.

From the list of unknowns, all you are left with to solve is the time for B. Using y = 0.5at^2, you should obtain approximately 7 seconds for your answer. The answer choice is A.

This might be a lengthy way of solving the problem, but I used this method to illustrate all the important points you should consider when solving a problem. Through practice, you will simplify the procedure. Hope this helps! Try #2 with the procedure, tweak it appropriately, and see what you get. Good .

 

[gridiron]

If you take any object at rest and it explodes, why is total momentum zero immediately after the explosion? I know P=mv

My line of thinking is this: Each fragment has a mass and a (very high) velocity after the explosion, so how can it not have a momentum?????

Anyone know why?

Hey! The reason the total momentum is zero is because the explosion occurred in a isolated system. An isolated system is free of external forces so the change in momentum will be equal to zero. Also remember that momemtum is a vector. Hope this helps and good .

 

[MSTPbound]

Hey! The reason the total momentum is zero is because the explosion occurred in a isolated system. An isolated system is free of external forces so the change in momentum will be equal to zero. Also remember that momemtum is a vector. Hope this helps and good .

Right. I just want to elaborate on this a little Playmakur, to address your visualization... Remember that Momentum is ALWAYS, ALWAYS, ALWAYS conserved. Initial Momentum will always be equal to final momentum. If the initial momentum is zero, then the final momentum MUST be zero. Want to visualize it?

Remember that when the object EXPLODES, Its fragments will be catapulted from each other in every possible direction, so all vector quantities in any direction are directly canceled by equal vector quantities in an opposite direction.

Does this make sense?

Good luck!

 

[Playmakur42]

Its fragments will be catapulted from each other in every possible direction, so all vector quantities in any direction are directly canceled by equal vector quantities in an opposite direction.

Does this make sense?

Yes, it does now. Thanks alot for taking the time to describe it more thoroughly. Very helpful.

When the last poster mentioned momentum was a VECTOR quantity, I had one of those "AHHHH" moments and understood how the momentum of all the fragments in the explosion summed to be zero and the initial momentum of 0 was conserved.

 

[BrokenGlass]

A particle moving at 10 m/s reverses its direction to move at 20 m/s in the opposite direction. If its acceleration is -10 m/s^2. what is the total distance that it travels...

I just dont understand the question being asked..can someone please clarify..this is # 61 from EK~
Thanks!

Given:

V0 = +10 m/s
Vf = -20 m/s
a = -10 m/s

This problem is about distance, NOT displacement. So it's probably best to to break it up into 2 parts:

part 1: v goes from 10 m/s to 0 m/s
part 2: v goes from 0 m/s to -20 m/s

Now lets tackle each part separately:

Part 1:

v = v0 + at
0 = 10 - 10*t
t = 1

x = x0 + v0 * t + 1/2 * a * t^2
x = 0 + 10 * 1 - 1/2 * 10 * 1^2 = 10 -5 = 5m

We started at zero m and ended up at 5m, so total distance covered in part 1 is 5 - 0 = 5 m

Part 2:

v = v0 + at
-20 = 0 - 10*t
t = 2

x = x0 + v0 * t + 1/2 * a * t^2
x = 5 + 0 - 1/2*10*2^2 = 5 - 20 = -15

We started at 5 m and ended up at -15m, so total distance covered in part 2 is 5 - (-15) = 20 m

And now for the death blow (to borrow the words of my friend before he beats me in Air Hockey): 5m + 20m = 25m

 

[BrokenGlass]

Right. I just want to elaborate on this a little Playmakur, to address your visualization... Remember that Momentum is ALWAYS, ALWAYS, ALWAYS conserved. Initial Momentum will always be equal to final momentum. If the initial momentum is zero, then the final momentum MUST be zero. Want to visualize it?

Remember that when the object EXPLODES, Its fragments will be catapulted from each other in every possible direction, so all vector quantities in any direction are directly canceled by equal vector quantities in an opposite direction.

Does this make sense?

Good luck!

Momentum is NOT always conserved. Momentum is conserved ONLY IF there is no NET external force acting on the system.

 

[BrokenGlass]

I am having a difficult time with Newton's Law of Universal Gravitation. Is there any one that can help me figure out this problem: I understand the problem but need to understand how to use the ratios that are involved. Can someone show me as it applies to these two problems.

1. Planet B has twice the mass of Planet A. Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10 m above the surface of Planet B and at the same time a 10 kg mass is dropped 10m above the surface of Planet A. If the mass on Planet B strikes the ground in 10 sec,the mass on Planet A strikes the ground in approx:
a.7, b. 10, c. 14, d.20 ans is a.
2. Planet A and B have the same mass . Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10m above the surface of Planet A and at the same a 5kg mass is dropped 10m above the surface of Planet B. If the mass on Planet B strikes the ground in 10 sec, the mass of Planet A strikes the ground in :
a. 2.5s b.5s c.10s d. 20s ans is b.

thanks

Ma = mass of planet a
Mb = mass of planet b
Ra = radius of planet a
Rb = radius of planet b



F = (G * M * m) / R^2
F = m*a
a = F/m
a = (G * M) / R^2

1)

Mb = 2*Ma
Rb = 2*Ra

In the formula a = (G * M) / R^2 if we double the mass and double the radius, we decrease the acceleration by a factor of 2.

=> acceleratino on planet b = 1/2 acceleration on planet a
=> acceleration on planet a = 2 * acceleartion on planet b

So time on planet a < time on planet b, so 7 sec is the right answer (no need to do any more work, we can stop right here).

2)

Mb = Ma
Rb = 2*Ra

=> acceleratino on planet b = 1/4 acceleration on planet a
=> acceleartion on planet a = 4 * acceleration on planet b

x = x0 + v0*t + 1/2 * a * t^2

x = 1/2 * a * t^2 (because initial displacement and initial velocity are both zero)

t^2 = (2*x)/a

t = sqrt((2*x)/a), where sqrt stands for square root

t ~ sqrt(1/a)

In the formula t ~ sqrt(1/a) if acceleration goes up by a factor of 4, t goes down by a factor of sqrt(4) = 2

10sec /2 = 5 sec

 

[Dwing]

A 250g ball is thrown straight into the air with a speed of 20m/s. If it flies to a height of 17m, find the work down by air resistance?

I just got into Energy Theorem but i dont get how to use the formula. can someone show me with using this question?
Thanks

 

[rcd]

A 250g ball is thrown straight into the air with a speed of 20m/s. If it flies to a height of 17m, find the work down by air resistance?

I just got into Energy Theorem but i dont get how to use the formula. can someone show me with using this question?
Thanks

first, figure out the height it would be w/ no air resistance:

v^2 = v0^2 + 2ay
0 = 20^2 + 2(-10)y
0 = 400 - 20y
-400 = -20y
y = 20m

then figure out the TotalEnergy w/out air resistance
TE = KE + PE
= 0 + mgy
TE = 20mg

now figure out TE w/ air resistance
TE = KE + PE
= 0 + mgy
= 17mg

now figure out the difference by conserving energy @ the top of the projectory:

TE no air resistance = TE air resistance + air resistance energy
20mg = 17mg + air
3mg = air
3(.250kg)10 = 7.50 J

 

[rcd]

first, figure out the height it would be w/ no air resistance:

v^2 = v0^2 + 2ay
0 = 20^2 + 2(-10)y
0 = 400 - 20y
-400 = -20y
y = 20m

then figure out the TotalEnergy w/out air resistance
TE = KE + PE
= 0 + mgy
TE = 20mg

now figure out TE w/ air resistance
TE = KE + PE
= 0 + mgy
= 17mg

now figure out the difference by conserving energy @ the top of the projectory:

TE no air resistance = TE air resistance + air resistance energy
20mg = 17mg + air
3mg = air
3(.250kg)10 = 7.50 J

Another way to do this:

Energy in the universe (energy of everything applicable to the problem) is neither created or destroyed:

1: E_ball[final] + E_air[final] = E_ball[initial] + E_air[initial]


E_ball[final] = KE + PE = 0 + mgy = 17mg
E_air[final] is the unknown

E_ball[initial] = KE + PE = .5 * mv^2 + 0 = .5m20^2 = 200m
E_air[initial] = 0

so, plugging into equation 1:
g=10
250 g = .250 kg

17mg + x = 200m
170m + x = 200m
x = 30m
x = 30 * .250
x = 7.5J

also, you may have to flip it to negative depending on the problem's definition of work:
either deltaE = q + w or deltaE = q - w

deltaE is positive, q is 0, so w would have to be positive for air in the first
deltaE is positive, q is 0, so w would have to be negative for air in the second

 

[jade1013]

I'm not so great at torque and I was working through question 60 in lecture 3 of the exam krackers book for physics. The answer says that the clockwise torque is the weight of the board times 0.3m. Where is the 0.3m coming from? Please help me!! Thank you!

 

[BrokenGlass]

I'm not so great at torque and I was working through question 60 in lecture 3 of the exam krackers book for physics. The answer says that the clockwise torque is the weight of the board times 0.3m. Where is the 0.3m coming from? Please help me!! Thank you!

Please post the entire problem statement, and the answer key.

 

[swifteagle43]

A crane lifts a 1000-kg steel beam off the ground, and sets it down on scaffolding 100 meters off the ground. All of the following are true EXCEPT:

A. The net work done on the steel beam is 9.8 × 105 J.
B. The net work done on the steel beam is 0 J.
C. The magnitude of the work done on the steel beam by gravity is 9.8 × 105 J.
D. The magnitude of the work done on the steel beam by the crane is 9.8 × 105 J.


Answer: A
Scanning the answer choices, you should notice that choice (A) and (B) cannot both be true, so the correct answer must be one of these. They both ask about the net work done on the steel beam. Remember the Work-Kinetic Energy Theorem, which states that the net work done on a system equals the change in kinetic energy of the system. Since the beam begins and ends with zero velocity, the change in kinetic energy is zero, and thus the net work is zero. Choice (A) is false, so this is the correct answer.

(B) Opposite. This statement is true. The beam has &#916;KE = 0, which means Wnet = 0.

(C) Opposite. This statement is true. The work done on the beam by gravity is negative (the beam does work on the Earth), because the direction of the force (downwards) is opposite the direction of displacement (upwards). Since work equals force times distance, the magnitude of the work is mgh = (1000 kg)(9.8 m/s2)(100 m) = 9.8 × 105 J.

(D) Opposite. This statement is true. The crane must do the same exact magnitude (but opposite sign) of work as gravity, because the net work done on the beam is zero. Again, work equals force times distance, so gravity does mgh = 9.5 × 105 J of work; this is the magnitude of work done by the crane.


Why is there no net work done on the block if it is raised 100 meters?

 

[cheezer]

The answer to your question lies in C and D

Same magnitude, but like you said W by gravity is negative (W=Fdcos180). W by crane is positive (W=Fdcos0). Add'em together and they cancel out.

Or you can look at it this way. Since net force is 0 given Fgravity and Fcranes opposite directions but equal values, then Net Work = 0 (d) = 0.

Hope that helped.

 

[Shrike]

Quick, qualitative approach to the crane problem: work is the tendency of a force to speed something up (more technically, to increase its kinetic energy). the beam didn't speed up or slow down: it went from rest to rest. hence, no net work done.

 

[swifteagle43]

Quick, qualitative approach to the crane problem: work is the tendency of a force to speed something up (more technically, to increase its kinetic energy). the beam didn't speed up or slow down: it went from rest to rest. hence, no net work done.

I thought work equaled Force * distance and a force of m * g is applied over a 100 meter length...

Why am I wrong? Or where is the flaw in this way of thinking?

 

[NontradICUdoc]

I thought work equaled Force * distance and a force of m * g is applied over a 100 meter length...

Why am I wrong? Or where is the flaw in this way of thinking?

You are partially correct. Work=Fd(cos theta)

 

[BrokenGlass]

A 250g ball is thrown straight into the air with a speed of 20m/s. If it flies to a height of 17m, find the work down by air resistance?

I just got into Energy Theorem but i dont get how to use the formula. can someone show me with using this question?
Thanks

If there were no friction:

E initial = E final, or KE = PE (since in the beginning all energy is kinetic and at the max height all energy is potential.

Note: since energy is a state function, we are free to choose any reference point we wish, so I assigned a value of
zero to internal energy at the point in time when the object is launched up.


With friction:

KE - work done by friction = PE

1/2 * m * v ^2 - work done by friction = m * g * h

250g = 1/4 kg

(1/2)*(1/4)*(20^2) - work done by friction = (1/4)*10*17

400/8 - 340/8 = work done by friciton

work done by friction = 60/8 = 15/2 = 7.5J

Friction converts 7.5J of mechanical energy to internal energy (mechanical energy is KE + PE).

 

[BrokenGlass]

Quick, qualitative approach to the crane problem: work is the tendency of a force to speed something up (more technically, to increase its kinetic energy). the beam didn't speed up or slow down: it went from rest to rest. hence, no net work done.

Shrike, how come you don't post more often? I always learn by reading your comments. Post more!

 

[HristosKaran]

An older gentleman drags a wooden crate (m=10 kg) containing his childhood toys out of storage. The 120 N force F that he applies on the crate makes an angle of 30° with the horizontal. The coefficient of kinetic friction &#956;k between the well-polished floor and wooden crate is experimentally determined to be 0.1. What is the frictional force f on the crate? (cos 30° = .866, sin 30° = .5)
A) 2.4 N
B) 3.8 N
C) 7.2 N
D) 9.8 N

Hey guys, the answer is C....But I can't seem to get these kinematic problems down..can someone help me out? Thanks

What I was going to do is F=N - mg sin Theta..but I know I'm missing a key point..

 

[BrokenGlass]

An older gentleman drags a wooden crate (m=10 kg) containing his childhood toys out of storage. The 120 N force F that he applies on the crate makes an angle of 30&#176; with the horizontal. The coefficient of kinetic friction &#956;k between the well-polished floor and wooden crate is experimentally determined to be 0.1. What is the frictional force f on the crate? (cos 30&#176; = .866, sin 30&#176; = .5)
A) 2.4 N
B) 3.8 N
C) 7.2 N
D) 9.8 N

Hey guys, the answer is C....But I can't seem to get these kinematic problems down..can someone help me out? Thanks

What I was going to do is F=N - mg sin Theta..but I know I'm missing a key point..

f_k = &#956;_k * N (by definition)

So in order to find f_k we need to figure out N (the normal force)

Forces in the Y direction cancel each other (i.e. sum of forces up = sum of forces down).

120 * sin 30 + N = m*g

N = m*g - 120 * sin 30 = 10 * 9.8 - 120 * 0.5 = 98 - 60 = 38 N

f_k = 0.1 * 38N = 3.8N

(We could have used 10 for g and the answer would have been 4N).

Anyhow, are you sure the anwer is C? I am gettin 3.8N, which is answer choice B.

 

[cheezer]

An older gentleman drags a wooden crate (m=10 kg) containing his childhood toys out of storage. The 120 N force F that he applies on the crate makes an angle of 30° with the horizontal. The coefficient of kinetic friction &#956;k between the well-polished floor and wooden crate is experimentally determined to be 0.1. What is the frictional force f on the crate? (cos 30° = .866, sin 30° = .5)
A) 2.4 N
B) 3.8 N
C) 7.2 N
D) 9.8 N

Hey guys, the answer is C....But I can't seem to get these kinematic problems down..can someone help me out? Thanks

What I was going to do is F=N - mg sin Theta..but I know I'm missing a key point..

I'd like to know too, I'm getting the same answer as BrokenGlass.

 

[HristosKaran]

f_k = &#956;_k * N (by definition)

So in order to find f_k we need to figure out N (the normal force)

Forces in the Y direction cancel each other (i.e. sum of forces up = sum of forces down).

120 * sin 30 + N = m*g

N = m*g - 120 * sin 30 = 10 * 9.8 - 120 * 0.5 = 98 - 60 = 38 N

f_k = 0.1 * 38N = 3.8N

(We could have used 10 for g and the answer would have been 4N).

Anyhow, are you sure the anwer is C? I am gettin 3.8N, which is answer choice B.

Yeah I went back and looked and the answer is B, my fault. Thats pretty much the same answer kaplan gave..my main problem is how did you know to add the Normal force to the force exerted? For some reason my intuition is telling me to subtract the normal force from mg since they oppose each other...Thanks

 

[BrokenGlass]

Yeah I went back and looked and the answer is B, my fault. Thats pretty much the same answer kaplan gave..my main problem is how did you know to add the Normal force to the force exerted? For some reason my intuition is telling me to subtract the normal force from mg since they oppose each other...Thanks

If you resolve the force exerted by the old man into X and Y components, the Y component points in the same directions as the Normal Force (i.e. up). Draw a diagram for these types of problems. Eventually, you'll be able to do them in your head...

 

[HristosKaran]

If you resolve the force exerted by the old man into X and Y components, the Y component points in the same directions as the Normal Force (i.e. up). Draw a diagram for these types of problems. Eventually, you'll be able to do them in your head...

That actually makes perfect sense thanks! Its always something small that throws me for a loop...BrokenGlass how prevalent are kinematic questions on the MCAT? Are they usually a couple discretes or a passage worth of questions?

 

[BrokenGlass]

That actually makes perfect sense thanks! Its always something small that throws me for a loop...BrokenGlass how prevalent are kinematic questions on the MCAT? Are they usually a couple discretes or a passage worth of questions?

I haven't take the MCAT yet, but you could get an entire passage on kinematics questions. btw, I think the term kinematics is referring to study of motion without worrying about what causes motion. So for instance, v = v0 + at is an equation from kinematics. The problem about the old man is not a kinematics problem because it invovles force analysis....

 

[gridiron]

Yeah I went back and looked and the answer is B, my fault. Thats pretty much the same answer kaplan gave..my main problem is how did you know to add the Normal force to the force exerted? For some reason my intuition is telling me to subtract the normal force from mg since they oppose each other...Thanks

Another way to look at the original problem: force is a tendency or influence on a object to accelerate it. Since the force is applied at an angle to the horizontal, acceleration will happen in which direction? Or, you can say that the force is being transferred to the object by the man in order to move it. In this case, work is being done. Which direction is the object going to move? This way you know which forces to consider when solving the problem or in finding the resultant forces on the object. What is also important is that force is a vector. Therefore, when drawing the free body diagram, you have to assign both a direction and magnitude. This means you have to consider which direction will be positive and negative but make sure to be consistent with all your sign convention.

 

[rcd]

A solid will buckle under pressures above 1.2E6 Pa. Its density is 4000kg/m^3. What is its maximum height of a circular column of the solid built on the Earth's surface? Note: 1atm = 1E5Pa.

So, EK says it's 30m (and they don't account for atmospheric pressure), I say it's 24m.


Why don't you need to take into account atmospheric pressure?

 

[BrokenGlass]

A solid will buckle under pressures above 1.2E6 Pa. Its density is 4000kg/m^3. What is its maximum height of a circular column of the solid built on the Earth's surface? Note: 1atm = 1E5Pa.

So, EK says it's 30m (and they don't account for atmospheric pressure), I say it's 24m.


Why don't you need to take into account atmospheric pressure?

P = F/A = mg/A = rho*V*g/A = rho*g*y (since V/A = y, where y is depth)



1.2 * 10^6 = 4000 * 10 * y

y = 30 m

The limiting pressure is already expressed relative to atmospheric pressure (that's how it's usually measured). Also, atmospheric pressure is an order of magnitude (about 10 times) smaller than what we are dealing with here.

 

[cheezer]

Damn BrokenGlass, you got your mechanics on lockdown

 

[BrokenGlass]

Damn BrokenGlass, you got your mechanics on lockdown

I wish. I am still working on it ...

 

[BrokenGlass]

Quick, qualitative approach to the crane problem: work is the tendency of a force to speed something up (more technically, to increase its kinetic energy). the beam didn't speed up or slow down: it went from rest to rest. hence, no net work done.

Shrike, I understand the answer key and yet it bothers me in the sense that the beam has more energy after it's lifted (in the form of potential energy). Where did this energy come from if net work is zero (and of course there is no heat transfer in this problem)?

&#916; E = &#916; KE + &#916; PE + &#916; Ei
&#916; E = W + q = W + 0 = W
W = &#916; KE + &#916; PE + &#916; Ei = &#916; KE + &#916; PE (since there is no change in internal energy &#916;Ei = 0)

So why doesn't the solution consider potential energy as well as KE?

What exactly is the definition of net work?

 

[xlr8]

Hi, I have a question on air resistance:

EK states that "Mass doesn't change the force of air resistance, but changes the path of the projectile experiencing the air resistance"

Then it goes on to ask a question where two balls (same size) of different mass X (has greater mass) and Y are dropped from a tower and reach terminal velocity.

During free fall there is constant velocity (no accel.) therefore the air resistance must counter the weight (mg) to allow constant velocity for each ball. However this would imply that the air resistant forces for each of the balls are unequal, contradicting their initial statement?

The answer to the problem is that Ball X has a greater velocity than Ball Y.
(Wouldn't this mean that Ball X has a greater air resistance than Ball Y?)

If someone could explain the flaws in my line of reasoning as well as why this is the answer it would be much appreciated, gracias.

 

[BrokenGlass]

Hi, I have a question on air resistance:

EK states that "Mass doesn't change the force of air resistance, but changes the path of the projectile experiencing the air resistance"

Then it goes on to ask a question where two balls (same size) of different mass X (has greater mass) and Y are dropped from a tower and reach terminal velocity.

During free fall there is constant velocity (no accel.) therefore the air resistance must counter the weight (mg) to allow constant velocity for each ball. However this would imply that the air resistant forces for each of the balls are unequal, contradicting their initial statement?

The answer to the problem is that Ball X has a greater velocity than Ball Y.
(Wouldn't this mean that Ball X has a greater air resistance than Ball Y?)

If someone could explain the flaws in my line of reasoning as well as why this is the answer it would be much appreciated, gracias.

Mass doesn't affect the force of air resistance. It affects acceleration. a = F/m. The heavier
ball has more inertia and air resistance will have less of an effect on the change in its velocity.

Also the force of air resistance is the same for these 2 balls only if they are travelling at the same velocity. When their velocities start to differ (which they will due to different accelerations), they will experience different force of air resistance.

HTH

 

[xlr8]

Mass doesn't affect the force of air resistance. It affects acceleration. a = F/m. The heavier
ball has more inertia and air resistance will have less of an effect on the change in its velocity.

Also the force of air resistance is the same for these 2 balls only if they are travelling at the same velocity. When their velocities start to differ (which they will due to different accelerations), they will experience different force of air resistance.

HTH

Ah, there is the missing information I needed. I was under the impression that the force of air resistance was the same throughout the entire fall, which really baffled me. So this is what I've understood:

Force remains constant in air resistance with a change of mass, just the acceleration is effected, which in turn takes its effect on the change in velocity:

So is it correct to picture it as the two balls falling, and their respective net forces on each are (Mg-F(air)), but since the Mg for the less massive ball is smaller, it will reach its terminal velocity quicker, and the larger ball will continue accelerating until it reaches its terminal velocity with a larger F(air), giving a larger velocity?

 

[BrokenGlass]

Ah, there is the missing information I needed. I was under the impression that the force of air resistance was the same throughout the entire fall, which really baffled me. So this is what I've understood:

Force remains constant in air resistance with a change of mass, just the acceleration is effected, which in turn takes its effect on the change in velocity:

So is it correct to picture it as the two balls falling, and their respective net forces on each are (Mg-F(air)), but since the Mg for the less massive ball is smaller, it will reach its terminal velocity quicker, and the larger ball will continue accelerating until it reaches its terminal velocity with a larger F(air), giving a larger velocity?

Yeah, I think that's correct.

 

[rcd]

A crane lifts a 1000-kg steel beam off the ground, and sets it down on scaffolding 100 meters off the ground. All of the following are true EXCEPT:

A. The net work done on the steel beam is 9.8 &#215; 105 J.
B. The net work done on the steel beam is 0 J.
C. The magnitude of the work done on the steel beam by gravity is 9.8 &#215; 105 J.
D. The magnitude of the work done on the steel beam by the crane is 9.8 &#215; 105 J.


Answer: A
Scanning the answer choices, you should notice that choice (A) and (B) cannot both be true, so the correct answer must be one of these. They both ask about the net work done on the steel beam. Remember the Work-Kinetic Energy Theorem, which states that the net work done on a system equals the change in kinetic energy of the system. Since the beam begins and ends with zero velocity, the change in kinetic energy is zero, and thus the net work is zero. Choice (A) is false, so this is the correct answer.

(B) Opposite. This statement is true. The beam has &#916;KE = 0, which means Wnet = 0.

(C) Opposite. This statement is true. The work done on the beam by gravity is negative (the beam does work on the Earth), because the direction of the force (downwards) is opposite the direction of displacement (upwards). Since work equals force times distance, the magnitude of the work is mgh = (1000 kg)(9.8 m/s2)(100 m) = 9.8 &#215; 105 J.

(D) Opposite. This statement is true. The crane must do the same exact magnitude (but opposite sign) of work as gravity, because the net work done on the beam is zero. Again, work equals force times distance, so gravity does mgh = 9.5 &#215; 105 J of work; this is the magnitude of work done by the crane.


Why is there no net work done on the block if it is raised 100 meters?

What a question!

Here's how to solve it:
1. Assume gravity is not in your system. Therefore, gravity does 0 work (it's a conservative force), and the crane does mgy work. The net change in energy of beam is mgy, so choice B would be the answer. But, choice C would be the answer too (gravity actually does 0 work)!

2. So, you must assume gravity is in the system. That means it is non-conservative. Which means, the beam had 0 PE to begin with, and 0 PE to end with.

Don't expect anything like this to be on the MCAT.

 

[gridiron]

Ah, there is the missing information I needed. I was under the impression that the force of air resistance was the same throughout the entire fall, which really baffled me. So this is what I've understood:

Force remains constant in air resistance with a change of mass, just the acceleration is effected, which in turn takes its effect on the change in velocity:

So is it correct to picture it as the two balls falling, and their respective net forces on each are (Mg-F(air)), but since the Mg for the less massive ball is smaller, it will reach its terminal velocity quicker, and the larger ball will continue accelerating until it reaches its terminal velocity with a larger F(air), giving a larger velocity?

By definition, terminal velocity is the velocity attained by an object when the drag force acting on the object balances (equals) the weight of the object minus the bouyant force acting on the object. This means that acceleration is equal to zero and the speed of the object is constant--this doesn't necessarily mean the velocity is constant since it is a vector quantity. The terminal velocity for an object is dependent on the ratio of the drag force acting on the object to the mass of the object. A higher drag force acting on the object will mean a smaller terminal velocity. Conversely, a higher mass will mean a greater terminal velocity.

 

[xlr8]

So the magnitude is constant, from the balancing of the forces, but the direction is derived the drag force for terminal velocity?



New question (on work):

Two objects A and B are placed on a spring, mass A has twice the mass of B. If the spring is depressed and released, object A will:

ExamKrackers states that they will both rise to the same height, but I'm having trouble making this connection. They state that mass is not proportional to height, but won't the elastic potential energy from the spring be converted into GPE (mgh), which will compensate heights due to the masses for each?

 

[gridiron]

The magnitude of the terminal velocity depends on the object. The terminal velocity will be greater for a heavier object then for a light object. For a falling body in air, terminal velocity will occur when the drag force due to air balances the weight of the body. Consider this example: you drop a rock and feather in air. Which is lighter? The feather is lighter. For the feather, the drag force will balance the weight of the feather more quickly then for the rock. This means the rock will accelerate for a longer period of time before it will experience terminal velocity. If you draw from the kinematic equations, acceleration is directly proportional to velocity squared, provided the initial velocity is zero. This is why the rock will experience a greater terminal velocity--it accelerates for a longer period of time.

Some important points:
1. Terminal velocity only happens when there is zero acceleration. What does this mean? This means the net force acting on the body is zero. For a falling body in air, this will happen when the drag force balances the weight of the body. The drag force is a result of the molecules of air colliding with the body to oppose the force of gravity.

2. The terminal velocity depends on the mass of the body--or the weight of the body. For heavier objects, it takes a longer time for the drag force to balance the weight of the object. Thus, the heavier object will accelerate for a longer period of time and will have the greater terminal velocity.

3. When terminal velocity is attained, the speed of the object remains constant.

 

[xlr8]

The magnitude of the terminal velocity depends on the object. The terminal velocity will be greater for a heavier object then for a light object. For a falling body in air, terminal velocity will occur when the drag force due to air balances the weight of the body. Consider this example: you drop a rock and feather in air. Which is lighter? The feather is lighter. For the feather, the drag force will balance the weight of the feather more quickly then for the rock. This means the rock will accelerate for a longer period of time before it will experience terminal velocity. If you draw from the kinematic equations, acceleration is directly proportional to velocity squared, provided the initial velocity is zero. This is why the rock will experience a greater terminal velocity--it accelerates for a longer period of time.

Some important points:
1. Terminal velocity only happens when there is zero acceleration. What does this mean? This means the net force acting on the body is zero. For a falling body in air, this will happen when the drag force balances the weight of the body. The drag force is a result of the molecules of air colliding with the body to oppose the force of gravity.

2. The terminal velocity depends on the mass of the body--or the weight of the body. For heavier objects, it takes a longer time for the drag force to balance the weight of the object. Thus, the heavier object will accelerate for a longer period of time and will have the greater terminal velocity.

3. When terminal velocity is attained, the speed of the object remains constant.

Thank you, all of that is very understandable now, but specifically which part of my example was wrong, where i said the velocity would remain constant?

 

[gridiron]

Thank you, all of that is very understandable now, but specifically which part of my example was wrong, where i said the velocity would remain constant?

You were correct that the velocity is constant--the acceleration is zero because the net forces acting on the object is equal to zero. When I teach, i'm used to stressing that constant speed doesn't imply constant velocity. Why? Say that you are driving a car at 5 m/s and then encounter a sharp turn but continue around the turn at 5 m/s. Your speed was constant, but your velocity wasn't because velocity is a vector and you changed direction. I just wanted to explain the concept further so terminal velocity could be understood. I hope this helps and good .

 

[rcd]

2 Questions.

First, you hold a loop stationary within a magnetic field. You increase the magnetic field. How does the current change in the loop?

Something like this:
xxxxxxxx
xxx(x)xxx
xxxxxxxx



Second, you have a stationary magnet that can't rotate. You also have a nearby stationary magnet that can rotate. Does the rotating magnet line up with the stationary magnet's magnetic field, or against it?


Thanks!

 

[trustforward]

f_k = &#956;_k * N (by definition)

So in order to find f_k we need to figure out N (the normal force)

Forces in the Y direction cancel each other (i.e. sum of forces up = sum of forces down).

120 * sin 30 + N = m*g

N = m*g - 120 * sin 30 = 10 * 9.8 - 120 * 0.5 = 98 - 60 = 38 N

f_k = 0.1 * 38N = 3.8N

(We could have used 10 for g and the answer would have been 4N).

Anyhow, are you sure the anwer is C? I am gettin 3.8N, which is answer choice B.

you are very much correct.. this is from kaplan's Qbank, and I just happened to cross this Q last night. The answer IS C (not B).

 

[xlr8]

Two objects A and B are placed on a spring, mass A has twice the mass of B. If the spring is depressed and released, object A will:

ExamKrackers states that they will both rise to the same height, but I'm having trouble making this connection. They state that mass is not proportional to height, but won't the elastic potential energy from the spring be converted into GPE (mgh), which will compensate heights due to the masses for each?

 

[BrokenGlass]

2 Questions.

First, you hold a loop stationary within a magnetic field. You increase the magnetic field. How does the current change in the loop?

Something like this:
xxxxxxxx
xxx(x)xxx
xxxxxxxx



Second, you have a stationary magnet that can't rotate. You also have a nearby stationary magnet that can rotate. Does the rotating magnet line up with the stationary magnet's magnetic field, or against it?


Thanks!

The induced current will oppose changes in the magnetic flux, so the induced current will be counterclockwise (to produce dots to cancel out the extra x's).